Question

Consider a point $P(1,2,3)$, plane $\pi :x+y+z=11$ and the line $L:\frac{x+1}{1}=\frac{y-12}{-2}=\frac{z-7}{2}$ The foot of the $\perp$ drawn from the point P meet the plane $\pi$ at M, then co-ordinate of M is

• A. $\left(\frac{8}{3},\frac{-11}{3},\frac{14}{3}\right)$
• B. $\left(\frac{-8}{3},\frac{-11}{3},\frac{-14}{3}\right)$
• C. $\left(\frac{8}{3},\frac{11}{3},\frac{14}{3}\right)$
• D. None of these

Answer 1

The correct option is C $\left(\frac{8}{3},\frac{11}{3},\frac{14}{3}\right)$

$P=(1,2,3)\to$ Point

$\Rightarrow \pi :x+y+z=11$

$L:\frac{x+1}{1}=\frac{y-12}{-2}=\frac{z-7}{2}$

Foot of perpendicular to the plane :

$\Rightarrow \frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{l-z_1}{c}=\frac{-(a{x}_1+b{y}_1+c{z}_1)}{a^2+b^2+c^2}$

$\frac{h-1}{1}=\frac{k-2}{1}=\frac{l-3}{1}=\frac{1+2+3}{3}$

$h-1=k-2=l-3=-2$

$h=-1,k=0,l=1$

The required point is

$(h+\frac{11}{3},k+\frac{11}{3},l+\frac{11}{3})$

$(\frac{8}{3},\frac{11}{3},\frac{14}{3})$

Hence the answer is $(\frac{8}{3},\frac{11}{3},\frac{14}{3}).$